Question

Find the shortest distance from origin to the plane x -2y -2z = 3.

Answer 1

Shortest distance from the origin (0, 0) to the plane x- 2y - 2z = 3 is

\(d = \left| \frac{0-2\times 0 - 2\times 0 -3}{\sqrt{1^2 + (-2)^2 + (-2)^2}}\right| = \left|\frac{-3}{\sqrt{1 + 4 + 4}}\right|\)

\(= \frac 3{\sqrt 9} = \frac 33 = 1\) unit